Carbon has valence electronic configuration of $2s^2$ $2p^2$. Beryllium has configuration $2s^2$.So carbon should have less ionisation potential than Beryllium as it is easier to remove electrons from $2p$ subshell than fully filled $2s$ subshell.

But I found that in JD Lee it is mentioned that carbon has more I.P. than beryllium. How can it be explained?


IP trends, group II

Emperically speaking, Be is on the "metallic" side of the table. Lower IE than non-metals is obviously expected.

A quick glance at the $IE_1$ trends for period II tells us that like the OP noted, Be certainly has an abnormally high $IE_1$, deviating significantly from the expected plot, due to the fully filled configuration.

However, Carbon (Z=6) has a higher nuclear charge(=more protons) than Be (Z=4), a smaller radius and hence a stronger hold on its valence electrons.

Even Nitrogen displays an exception, due to half-filled $2p$ subshell, as evident in the plot.

Just because of a tiny bump in the plot, expecting N and Be to have higher IE than all the elements of the periodic table is clearly an exaggeration.

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These periodic trends in 1st ionisation energy (I.E.) are always the result of more than one factor changing. The most important factors are:

  1. The effective nuclear charge (Zeff) that the valence electron being removed experiences (roughly equal, for s and p blocks, to the number of valence electrons)
  2. The potential energy of the sublevel (1s, 2s, 2p etc) that the electron is being removed from
  3. Any electron-electron repulsion that the electron being removed experiences in its orbital enter image description here As we move left to right across the period, factor 1 increases and this dominates the overall shape shown in the graph. A drop occurs from Be to B as we move to the higher potential energy 2p orbital (so the electron is easier to remove) and this is more important than the increasing Zeff from 2+ to 3+. As we move to C, the Zeff increases to 4+, so it has a higher I.E. than B (same sublevel, 2p).

The reason that C's I.E. is higher than Be also is because the change in Zeff from Be to C is +2 to +4 (it roughly doubles) and this is more important than the difference in potential energy between the 2s and 2p sublevels, so factor 1 takes over again, over factor 2.

I haven't discussed factor 3. Anytime an electron is removed from a paired orbital, it is a little easier (lower I.E.) than exactly the same situation (same Zeff, same sublevel potential energy) from an unpaired orbital, due to electron-electron repulsion within the paired orbital. It is a small factor between Be and C (Be is a paired 2s, C is an unpaired 2p) but it is very minor compared to factors 1 and 2 changing between these species, so C is still harder to remove.

It is tempting to say that a filled sublevel, or orbital, is "stable" = higher I.E. This statement comes not from the fact that the sublevel is full (these paired electrons are actually easier to remove than you would expect, due to electron-electron repulsion) but because filled sublevels always correspond to the highest Zeff for that sublevel.

You can employ these arguments to explain all the trends in the simple (s and p block) 2nd and 3rd periods. Things get much more complicated in other parts of the table!

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